Problem: Recently Duff has been a soldier in the army. Malek is her commander.

Their country, Andarz Gu has n cities (numbered from 1 to n) and n - 1 bidirectional roads. Each road connects two different cities. There exist a unique path between any two cities.

There are also m people living in Andarz Gu (numbered from 1 to m). Each person has and ID number. ID number of i - th person is i and he/she lives in city number ci. Note that there may be more than one person in a city, also there may be no people living in the city.

<image>

Malek loves to order. That's why he asks Duff to answer to q queries. In each query, he gives her numbers v, u and a.

To answer a query:

Assume there are x people living in the cities lying on the path from city v to city u. Assume these people's IDs are p1, p2, ..., px in increasing order. 

If k = min(x, a), then Duff should tell Malek numbers k, p1, p2, ..., pk in this order. In the other words, Malek wants to know a minimums on that path (or less, if there are less than a people).

Duff is very busy at the moment, so she asked you to help her and answer the queries.

Input

The first line of input contains three integers, n, m and q (1 ≤ n, m, q ≤ 105).

The next n - 1 lines contain the roads. Each line contains two integers v and u, endpoints of a road (1 ≤ v, u ≤ n, v ≠ u).

Next line contains m integers c1, c2, ..., cm separated by spaces (1 ≤ ci ≤ n for each 1 ≤ i ≤ m).

Next q lines contain the queries. Each of them contains three integers, v, u and a (1 ≤ v, u ≤ n and 1 ≤ a ≤ 10).

Output

For each query, print numbers k, p1, p2, ..., pk separated by spaces in one line.

Examples

Input

5 4 5
1 3
1 2
1 4
4 5
2 1 4 3
4 5 6
1 5 2
5 5 10
2 3 3
5 3 1


Output

1 3
2 2 3
0
3 1 2 4
1 2

Note

Graph of Andarz Gu in the sample case is as follows (ID of people in each city are written next to them):

<image>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 5;
const int mod = 1e9 + 7;
const int inf = (1 << 30);
const int block = 316;
vector<int> minn[maxn][30], v[maxn];
int par[maxn][30], h[maxn];
vector<int> merge(vector<int> fi, vector<int> se) {
  int idx1 = 0, idx2 = 0, cnt = 0;
  vector<int> ret;
  while (cnt < 10 and (idx1 < fi.size() or idx2 < se.size())) {
    if (idx1 < fi.size()) {
      if (idx2 == se.size() or fi[idx1] < se[idx2])
        ret.push_back(fi[idx1++]);
      else
        ret.push_back(se[idx2++]);
    } else
      ret.push_back(se[idx2++]);
    cnt++;
  }
  return ret;
}
bool visited[maxn];
void dfs(int u, int parent = -1) {
  visited[u] = 1;
  if (parent != -1) h[u] = h[parent] + 1;
  par[u][0] = parent;
  for (int i = 1; i <= 20 and par[u][i - 1] != -1; i++) {
    par[u][i] = par[par[u][i - 1]][i - 1];
    minn[u][i] = merge(minn[u][i - 1], minn[par[u][i - 1]][i - 1]);
  }
  for (auto w : v[u])
    if (!visited[w]) dfs(w, u);
}
int lca(int fi, int se) {
  if (fi == se) return fi;
  if (h[fi] < h[se]) return lca(se, fi);
  for (int i = 20; i >= 0; i--)
    if (h[fi] - (1 << i) >= h[se]) fi = par[fi][i];
  if (fi == se) return fi;
  for (int i = 20; i >= 0; i--) {
    if (par[fi][i] != par[se][i]) {
      fi = par[fi][i];
      se = par[se][i];
    }
  }
  return par[fi][0];
}
void solve(int fi, int se, int LCA, int a) {
  vector<int> ans;
  for (int i = 20; i >= 0; i--) {
    if (h[fi] - (1 << i) >= h[LCA]) {
      for (auto w : minn[fi][i]) ans.push_back(w);
      fi = par[fi][i];
    }
    if (h[se] - (1 << i) >= h[LCA]) {
      for (auto w : minn[se][i]) ans.push_back(w);
      se = par[se][i];
    }
  }
  for (auto w : minn[LCA][0]) ans.push_back(w);
  sort(ans.begin(), ans.end());
  cout << min(a, (int)ans.size()) << " ";
  for (int i = 0; i < min(a, (int)ans.size()); i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  int n, m, q;
  cin >> n >> m >> q;
  for (int i = 1; i <= n - 1; i++) {
    int u, w;
    cin >> u >> w;
    v[u].push_back(w);
    v[w].push_back(u);
  }
  for (int i = 1; i <= m; i++) {
    int c;
    cin >> c;
    if (minn[c][0].size() < 10) minn[c][0].push_back(i);
  }
  dfs(1);
  for (int i = 1; i <= q; i++) {
    int u, w, a;
    cin >> u >> w >> a;
    int LCA = lca(u, w);
    solve(u, w, LCA, a);
  }
}